T1
思路
把题目读错了。P[i][j]单调不升我看成单调上升了23333
然后正解是忽略上面这句话？然后我就A了？？？
用f[i][j]表示前i场赢了j场的概率，那么将当前这一场赢或输分类dp就好了。
代码
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<ctime>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1010;
ll read() {
ll x = 0, f = 1;
char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
double P[N][N],f[N][N];
int main() {
freopen("game.in","r",stdin);
freopen("game.out","w",stdout);
int n = read();
for(int i = 1;i <= n;++i)
for(int j = 0;j < i; ++j)
scanf("%lf",&P[i][j]);
f[0][0] = 1;
for(int i = 1;i <= n;++i) {
for(int j = 0;j <= i;++j) {
if(j >= 1)
f[i][j] += f[i-1][j-1] * P[i][j-1];
if(j < i)
f[i][j] += f[i-1][j] * (1-P[i][j]);
}
}
double ans = 0;
for(int i = 0;i <= n;++i)
ans += f[n][i] * (double)i;
printf("%.2lf",ans);
return 0;
}
/*
2
0.5
0.5 0.5
*/
T2
思路
meet in the middle + 二分
先正着搜一半，将所有能够构成的价格存到一个数组里。然后排个序，再倒着搜一半。搜完之后二分一下剩下的价格找个最优值就好了。
代码
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<ctime>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 50;
ll read() {
ll x = 0, f = 1;
char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
ll a[N];
ll tmp[2000000 + 100];
int n,tot;
ll W;
ll ans;
void dfs1(int pos,ll now) {
if(now > W) return;
if(pos > n / 2) {
tmp[++tot] = now;
return;
}
dfs1(pos + 1,a[pos] + now);
dfs1(pos + 1,now);
}
ll erfen(int x) {
int l = 1,r = tot;
ll re = 0;
while(l <= r) {
int mid = (l + r) >> 1;
if(tmp[mid] <= x) re = tmp[mid],l = mid + 1;
else r = mid - 1;
}
return re;
}
void dfs2(int pos,ll now) {
if(now > W) return;
if(pos <= n / 2) {
ll x = W - now;
ans = min(ans,x - erfen(x));
return;
}
dfs2(pos - 1,now + a[pos]);
dfs2(pos - 1,now);
}
int main() {
freopen("cake.in","r",stdin);
freopen("cake.out","w",stdout);
n = read(),W = read();
ans = W;
for(int i = 1;i <= n;++i) a[i] = read();
dfs1(1,0);
sort(tmp + 1,tmp + tot + 1);
dfs2(n,0);
cout<<ans;
return 0;
}
/*
4 50
1 2 3 4
4 5
1 2 3 4
*/
T3
20分思路
对于前20分。直接枚举所有情况，然后暴力判断即可
30分思路
对于n=0的情况，显然每个点都有m种取值,输出(m^{(h*w)})就行了
50分思路
当n=1的时候，考虑给出的这个矩形中的点的最大值必须是w，所以用全部情况减去不满的情况即可。全部情况是(w^{siz}),siz是这个矩形的大小，满足的情况就是这个矩形中所有值都比w小。所以这个矩形内部的情况就有(w^{siz}-(w-1)^{siz})。然后再考虑矩形外部的情况。矩形外部当然还和上面一样，每个点都有m种取值，所以情况数量就是(m^{h*w-siz})将这个两者乘起来就行了。
100分思路
我也不知道为什么我的错误代码也a了2333。RP--。
代码(下方高能，慎看！！)
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<map>
#include<ctime>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 10010,mod = 1e9 + 7;
map<int,int>Max,May;
ll read() {
ll x = 0, f = 1;
char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
int h,w,n,m;
struct node {
int x_1,x_2,y_1,y_2,w,siz;
node() {
siz = 0;
}
}Q[N];
ll qm(ll x,int y) {
if(x == 0) return 0;
ll ans = 1;
for(;y;y >>=1 ,x = x * x % mod)
if(y & 1) ans = ans * x % mod;
return ans;
}
namespace BF1 {
int a[10][10];
ll ans = 0;
void check() {
for(int i = 1;i <= n;++i) {
int Max = -1;
int X1 = Q[i].x_1,X2 = Q[i].x_2,Y1 = Q[i].y_1,Y2 = Q[i].y_2;
for(int j = 1;j <= h;++j) {
for(int k = 1;k <= w;++k) {
if(j >= X1 && j <= X2 && k >= Y1 && k <= Y2) {
Max = max(Max,a[j][k]);
}
}
}
if(Max != Q[i].w) return;
}
ans++;
}
void dfs(int x,int y) {
if(x > h) {
check();
return;
}
for(int i = 1;i <= m;++i) {
a[x][y] = i;
if(y == w) dfs(x + 1,1);
else dfs(x,y+1);
}
}
void Main() {
for(int i = 1;i <= n; ++i) {
Q[i].x_1 = read(),Q[i].y_1 = read(),Q[i].x_2 = read(),Q[i].y_2 = read(),Q[i].w = read();
}
dfs(1,1);
cout<<ans;
}
}
namespace BF2 {
void Main() {
for(int i = 1;i <= n; ++i) {
Q[i].x_1 = read(),Q[i].y_1 = read(),Q[i].x_2 = read(),Q[i].y_2 = read(),Q[i].w = read();
}
ll siz = (Q[1].x_2 - Q[1].x_1 + 1) * (Q[1].y_2 - Q[1].y_1 + 1);
ll ans = 0;
ans = (qm(Q[1].w,siz) - qm(Q[1].w - 1,siz) + mod) % mod;
cout<<ans * qm(m,h * w - siz) % mod;
return;
}
}
namespace BF3 {
int Wx[N * 10],Wy[N * 10],xjs = 0,yjs = 0,tmpx[N * 10],tmpy[N * 10];
int NEWH = 0,NEWW = 0;
ll SIZ;
void init() {
for(int i = 1;i <= n;++i) {
Q[i].x_1 = tmpx[++xjs] = read();Q[i].y_1 = tmpy[++yjs] = read();
Q[i].x_2 = tmpx[++xjs] = read();Q[i].y_2 = tmpy[++yjs] = read();
Q[i].w = read();
tmpx[++xjs] = Q[i].x_1 - 1;
tmpy[++yjs] = Q[i].y_1 - 1;
}
tmpx[++xjs] = 1,tmpx[++xjs] = h;
tmpy[++yjs] = 1,tmpy[++yjs] = w;
}
void lisan() {
sort(tmpx + 1,tmpx + xjs + 1);
sort(tmpy + 1,tmpy + yjs + 1);
int now = 0;
for(int i = 1;i <= xjs;++i) {
if(tmpx[i] == tmpx[i - 1]) continue;
Max[tmpx[i]] = ++now;
Wx[now] = tmpx[i] - tmpx[i - 1];
}
now = 0;
for(int i = 1; i <= yjs; ++i) {
if(tmpy[i] == tmpy[i - 1]) continue;
May[tmpy[i]] = ++now;
Wy[now] = tmpy[i] - tmpy[i-1];
}
for(int i = 1;i <= n;++i) {
Q[i].x_1 = Max[Q[i].x_1];Q[i].x_2 = Max[Q[i].x_2];
Q[i].y_1 = May[Q[i].y_1];Q[i].y_2 = May[Q[i].y_2];
}
NEWH = Max[h],NEWW = May[w];
}
ll calc(int x) {
ll siz = Q[x].siz;
ll re = (qm(Q[x].w,siz) - qm(Q[x].w - 1,siz) + mod) % mod;
return re;
}
void solve() {
for(int i = 1;i <= n;++i) {
int X1 = Q[i].x_1,X2 = Q[i].x_2,Y1 = Q[i].y_1,Y2 =Q[i].y_2;
// printf("%d %d %d %d
",X1,Y1,X2,Y2);
for(int j = 1;j <= NEWH; ++j) {
if(j < X1 || j > X2) continue;
for(int k = 1;k <= NEWW;++k) {
if(k < Y1 ||k > Y2) continue;
Q[i].siz += Wx[j] * Wy[k];
for(int z = 1; z <= n;++z) {
if(z==i) continue;
if(j >= Q[z].x_1 && j <= Q[z].x_2 && k >= Q[z].y_1 && k <= Q[z].y_2) {
if(Q[z].w > Q[i].w) continue;
if(Q[z].w == Q[i].w) {
if(z > i) {
Q[i].siz -= Wx[j] * Wy[k];
break;
}
}
if(Q[z].w < Q[i].w) {
Q[i].siz -= Wx[j] * Wy[k];
break;
}
}
}
}
}
}
ll ans = 1;
for(int i = 1;i <= n;++i) {
SIZ -= Q[i].siz;
ans *= calc(i);
ans %= mod;
// cerr<<Q[i].siz<<endl;
}
// cerr<<<<endl;
ans *= qm(m,SIZ);
ans %= mod;
cout<<ans;
}
void Main() {
SIZ = h * w;
init();
lisan();
// for(int i = 1;i <= h;++i) {
// printf("%d ",Wy[i]);
// }
solve();
}
}
int main() {
freopen("grid.in","r",stdin);
freopen("grid.out","w",stdout);
h = read(),w = read(),m = read(),n = read();
if(n == 0) {
cout<<qm(m,h * w);
return 0;
}
if(h <= 3 && w <= 3) {
BF1::Main();
return 0;
}
if(n == 1) {
BF2::Main();
return 0;
}
BF3::Main();
return 0;
}
/*
2 3 2 1
1 2 2 3 1
*/
总结
期望得分100 + 100 + 50 = 250
实际得分100 + 100 + 100 = 300
T3交卷之前拍出了错误，数据里没有？？？(雾)
一言
在一出戏锣鼓喧声的开幕时，就要知道散场后灯火尽消的凉清。 ——因为懂得，所以慈悲